访问方法内部的变量,并允许使用segue将其传递给其他视图控制器 [英] Accessing a variable inside a method and allowing it to be passed using segue to other view controller
本文介绍了访问方法内部的变量,并允许使用segue将其传递给其他视图控制器的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
func collectionView(_ collectionView: UICollectionView, cellForItemAt indexPath: IndexPath) -> UICollectionViewCell{
let cell = collectionView.dequeueReusableCell(withReuseIdentifier: cellIdentifier, for: indexPath) as! FeedCollectionViewCell
cell.postTextView.text = posts.reversed()[indexPath.row].caption
cell.postImageView.image = UIImage(named: "photoplaceholder.jpg")
cell.priceTextView.text = posts.reversed()[indexPath.row].price
cell.categoryTextView.text = posts.reversed()[indexPath.row].category
cell.usernameLabel.text = posts.reversed()[indexPath.row].username
cell.buttonEvents = {
let storyboard = UIStoryboard(name: "Main" , bundle: nil)
let chatViewController =
storyboard.instantiateViewController(withIdentifier: "chat")
self.present(chatViewController, animated: true,completion: nil)
var receiverIDNumber = cell.usernameLabel.text
}
我想将receiverIDNumber传递给另一个viewcontroller,但是由于该变量位于方法内部,所以我无法执行此操作.
I want to pass receiverIDNumber to another viewcontroller,however i am not able do as the variable is inside the method.
override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
let chatViewController = segue.destination as! ChatViewController
chatViewController.receivedString = receiverIDNumber
}
推荐答案
您可以尝试一下,因为 prepareForSegue
仅在 performSegue
不存在的情况下触发>
You can try this as prepareForSegue
is only triggered with performSegue
not present
let chatViewController = storyboard.instantiateViewController(withIdentifier: "chat") as! ChatViewController
chatViewController.receivedString = cell.usernameLabel.text
self.present(chatViewController, animated: true,completion: nil)
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